Simplify the following expression and state the condition under which the simplification is valid. You can assume that $p \neq 0$. $q = \dfrac{4p^2 + 28p}{8p} \times \dfrac{4}{4(p + 7)} $
Explanation: When multiplying fractions, we multiply the numerators and the denominators. $q = \dfrac{ (4p^2 + 28p) \times 4 } { 8p \times 4(p + 7) } $ $ q = \dfrac {4 \times 4p(p + 7)} {8p \times 4(p + 7)} $ $ q = \dfrac{16p(p + 7)}{32p(p + 7)} $ We can cancel the $p + 7$ so long as $p + 7 \neq 0$ Therefore $p \neq -7$ $q = \dfrac{16p \cancel{(p + 7})}{32p \cancel{(p + 7)}} = \dfrac{16p}{32p} = \dfrac{1}{2} $